📚 Year 10 CCEA Further Mathematics: Unit Test Mock Paper Walkthrough | Year 10 CCEA 进阶数学：单元测试模拟卷解析
Welcome to this detailed walkthrough of a mock unit test designed for Year 10 CCEA Further Mathematics. The paper covers the core topics of the syllabus, including algebra, functions, trigonometry, introductory calculus, vectors, matrices, indices and logarithmic equations. Every question is broken down step by step, with clear reasoning so that you can understand the method, master the key skill and avoid the mistakes that cost marks most often. Use this guide alongside your own revision to build both confidence and accuracy.
欢迎阅读这份为 Year 10 CCEA 进阶数学设计的单元测试模拟卷解析。本卷覆盖教学大纲的核心主题，包括代数、函数、三角学、微积分入门、向量、矩阵、指数与对数方程。每道题目都配有逐步解析与清晰推理，帮助你理解方法、掌握关键技能，并避开最常见的失分错误。请将这份解析与你的复习计划结合使用，以提升信心与解题准确度。
1. Paper Overview and Assessment Objectives | 试卷概览与考核目标
This mock unit test is written in the style of a CCEA Further Mathematics unit assessment. It is a calculator paper, lasts 1 hour 30 minutes and carries 100 marks. The questions are arranged roughly in order of difficulty, but the later topics are weighted more heavily, so do not spend too long on the first few questions.
这份模拟单元测试按照 CCEA 进阶数学单元测评的风格编写。它允许使用计算器，时长 1 小时 30 分钟，总分 100 分。题目大致按难度递增排列，但后段主题分值更重，因此不要在前几题上花费过多时间。
The assessment objectives are divided into three strands: recall of facts and definitions, routine procedural fluency, and problem solving in unfamiliar contexts. Roughly 40 per cent of the marks are for method, 50 per cent for accuracy, and 10 per cent for quality of written communication, which means clear setting out is genuinely worth marks.
考核目标分为三条主线：事实与定义的记忆、常规程序的熟练度，以及陌生情境下的问题解决。约 40% 的分值给方法，50% 给准确性，10% 给书面表达质量 —— 这意味着清晰的书写过程本身就是在拿分。
| Topic | Questions | Marks |
|---|---|---|
| Algebra and completing the square | 1 – 2 | 12 |
| The discriminant | 3 | 8 |
| Functions, composite and inverse | 4 | 10 |
| Trigonometry | 5 – 6 | 12 |
| Differentiation | 7 – 8 | 14 |
| Tangents, normals and stationary points | 9 | 10 |
| Vectors | 10 | 10 |
| Matrices | 11 | 10 |
| Logarithms and exponentials | 12 | 8 |
| Surds and indices | 13 | 6 |
The single most useful habit for this paper is to write down the formula or definition you are about to use before you substitute any numbers. Examiners award method marks for a correct starting statement even when the final answer is wrong.
应对这份试卷最有用的习惯是：在代入任何数字之前，先写下你要使用的公式或定义。即使最终答案错误，只要起始表述正确，考官依然会给方法分。
2. Completing the Square | 配方法
Question 1: solve the quadratic equation 2x² – 5x – 3 = 0 by completing the square. You must show every stage of the method.
第 1 题：用配方法解二次方程 2x² – 5x – 3 = 0，必须写出方法的每一个步骤。
Step 1: divide every term by 2 so that the coefficient of x² becomes 1. This gives x² – (5/2)x – 3/2 = 0.
步骤 1：将每一项除以 2，使 x² 的系数变为 1，得到 x² – (5/2)x – 3/2 = 0。
Step 2: move the constant term to the right-hand side, so x² – (5/2)x = 3/2.
步骤 2：把常数项移到等号右边，得到 x² – (5/2)x = 3/2。
cient of x, which is 5/4, then square it to obtain (5/4)² = 25/16. Add this to both sides.
步骤 3：取 x 的系数的一半，即 5/4，再平方得到 (5/4)² = 25/16，并将它加到等式两边。
x² – (5/2)x + 25/16 = 3/2 + 25/16
Step 4: the left-hand side is now a perfect square. Write it as (x – 5/4)². On the right, use a common denominator: 3/2 = 24/16, so 24/16 + 25/16 = 49/16.
步骤 4：左边现在是完全平方，写成 (x – 5/4)²。右边通分：3/2 = 24/16，于是 24/16 + 25/16 = 49/16。
(x – 5/4)² = 49/16
tep 5: take the square root of both sides. Remember that a square root has two signs, so x – 5/4 = ±√49/16 = ±7/4.
步骤 5：两边同时开平方。注意平方根有正负两个符号，所以 x – 5/4 = ±√49/16 = ±7/4。
Step 6: solve for x. This gives x = 5/4 ± 7/4, so x = (5 + 7)/4 = 12/4 = 3 and x = (5 – 7)/4 = –2/4 = –1/2.
步骤 6：解出 x。得到 x = 5/4 ± 7/4，因此 x = (5 + 7)/4 = 12/4 = 3，以及 x = (5 – 7)/4 = –2/4 = –1/2。
The solutions are x = 3 and x = –½. Check quickly: 2(3)² – 5(3) – 3 = 18 – 15 – 3 = 0, and 2(–½)² – 5(–½) – 3 = ½ + 5/2 – 3 = 0. Both roots are correct.
方程的解为 x = 3 和 x = –½。快速检验：2(3)² – 5(3) – 3 = 18 – 15 – 3 = 0，而 2(–½)² – 5(–½) – 3 = ½ + 5/2 – 3 = 0，两个根都正确。
The same completed-square form gives you the vertex of the curve: y = 2x² – 5x – 3 = 2(x – 5/4)² – 49/8, so the minimum point is (5/4, –49/8) and the axis of symmetry is x = 5/4, which is exactly the midpoint of the two roots.
同一个配方形式还能给出曲线的顶点：y = 2x² – 5x – 3 = 2(x – 5/4)² – 49/8，因此最低点为 (5/4, –49/8)，对称轴为 x = 5/4，正好是两个根的中点。
3. The Discriminant | 判别式
any quadratic equation ax² + bx + c = 0 with a ≠ 0, the discriminant is defined as Δ = b² – 4ac. It tells you the nature of the roots without solving the equation.
对任意二次方程 ax² + bx + c = 0（a ≠ 0），判别式定义为 Δ = b² – 4ac。它能在不解方程的情况下告诉你根的性质。
real root, and the curve touches the x-axis at a single point. If Δ < 0 there are no real roots, and the curve never meets the x-axis.
需要掌握三种情况：若 Δ > 0，有两个不同的实根；若 Δ = 0，有一个重根，曲线与 x 轴只有一个交点（相切）；若 Δ < 0，没有实根，曲线与 x 轴不相交。
Example 1: for 2x² – 5x – 3 = 0 we have a = 2, b = –5, c = –3, so Δ = (–5)² – 4(2)(–3) = 25 + 24 = 49.
例 1：对 2x² – 5x – 3 = 0，有 a = 2，b = –5，c = –3，所以 Δ = (–5)² – 4(2)(–3) = 25 + 24 = 49。
Because 49 > 0 the equation has two distinct real roots, and since 49 is a perfect square the roots are rational, which agrees with the exact answers 3 and –½ found by completing the square.
由于 49 > 0，方程有两个不同的实根；又因为 49 是完全平方数，所以根是有理数，这与用配方法得到的精确解 3 和 –½ 完全一致。
Example 2: 3x² – 4x + 5 = 0 gives Δ = (–4)² – 4(3)(5) = 16 – 60 = –. Since Δ < 0, there are no real roots and the parabola lies entirely above the x-axis.
例 2：对 3x² – 4x + 5 = 0，Δ = (–4)² – 4(3)(5) = 16 – 60 = –44。由于 Δ < 0，方程没有实根，抛物线整体位于 x 轴上方。
st: find the values of k for which x² + kx + 9 = 0 has equal roots. Setting Δ = 0 gives k² – 36 = 0, so k = 6 or k = –6. Always state both values, because the negative one is easy to lose.
例 3 是经典考法：求使 x² + kx + 9 = 0 有等根的 k 值。令 Δ = 0 得 k² – 36 = 0，所以 k = 6 或 k = –6。务必写出两个值，因为负值很容易被漏掉。
- If the question says 'touches the x-axis' or 'has a repeated root', use Δ = 0.
如果题目说"与 x 轴相切"或"有重根"，就用 Δ = 0。 - If it says 'cuts the x-axis twice', use Δ > 0.
如果说"与 x 轴有两个交点"，就用 Δ > 0。 - If it says 'does not meet the x-axis', use Δ < 0.
如果说"不与 x 轴相交"，就用 Δ < 0。
4. Functions: Domain, Range, Composite and Inverse | 函数：定义域、值域、复合与反函数
Question 4 concerns the functions f(x) = 2x + 3 and g(x) = x², both with domain x ∈ ℝ. Write down fg(x), gf(x) and f⁻¹(x), and explain why gf(x) is not the same as fg(x).
第 4 题涉及函数 f(x) = 2x + 3 与 g(x) = x²，
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